## MATH1111 Quizzes

Anti-differentiation (analytically) Quiz
Web resources available Questions

This quiz tests the work covered in the lecture on the analytical interpretation of the anti-derivative and corresponds to Section 6.2 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 2.

There are quizzes and web links on this topic at http://quiz.econ.usyd.edu.au/mathquiz/integration/index.php

Which of the following statements are correct? (Zero or more options can be correct)
 a) $\int {x}^{2}\phantom{\rule{0.3em}{0ex}}dx=2x+c$ b) $\int 2{x}^{3}\phantom{\rule{0.3em}{0ex}}dx=\frac{1}{2}{x}^{4}+c$ c) $\int \left({x}^{2}+3x+1\right)\phantom{\rule{0.3em}{0ex}}dx=\frac{1}{3}{x}^{3}+3{x}^{2}+x+c$ d) $\int sinx\phantom{\rule{0.3em}{0ex}}dx=cosx+c$ e) $\int cosx\phantom{\rule{0.3em}{0ex}}dx=sinx+c$

There is at least one mistake.
For example, choice (a) should be False.
$2x$ is the derivative of ${x}^{2}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (b) should be True.
$\frac{d}{dx}\left(\frac{1}{2}{x}^{4}+c\right)=\frac{4}{2}{x}^{3}=2{x}^{3}$
There is at least one mistake.
For example, choice (c) should be False.
$\frac{d}{dx}\left(\frac{1}{3}{x}^{3}+3{x}^{2}+x+c\right)={x}^{2}+6x+1$ so the statement is incorrect.
There is at least one mistake.
For example, choice (d) should be False.
$\frac{d}{dx}\left(cosx+c\right)=-sinx$ so the statement is incorrect.
There is at least one mistake.
For example, choice (e) should be True.
$\frac{d}{dx}\left(sinx+c\right)=cosx$
Correct!
1. False $2x$ is the derivative of ${x}^{2}\phantom{\rule{0.3em}{0ex}}.$
2. True $\frac{d}{dx}\left(\frac{1}{2}{x}^{4}+c\right)=\frac{4}{2}{x}^{3}=2{x}^{3}$
3. False $\frac{d}{dx}\left(\frac{1}{3}{x}^{3}+3{x}^{2}+x+c\right)={x}^{2}+6x+1$ so the statement is incorrect.
4. False $\frac{d}{dx}\left(cosx+c\right)=-sinx$ so the statement is incorrect.
5. True $\frac{d}{dx}\left(sinx+c\right)=cosx$
Which of the following correctly gives the anti-derivative $F\left(x\right)$ with ${F}^{\prime }\left(x\right)=2sinx+2{x}^{2}+5x+1$ where $F\left(0\right)=2$? Exactly one option must be correct)
 a) $F\left(x\right)=2cosx+{x}^{3}+\frac{5}{2}{x}^{2}+x$ b) $F\left(x\right)=-2cosx+\frac{1}{3}{x}^{3}+\frac{1}{2}{x}^{2}+x+4$ c) $F\left(x\right)=2cosx+\frac{2}{3}{x}^{3}+\frac{5}{2}{x}^{2}+x$ d) $F\left(x\right)=-2cosx+\frac{2}{3}{x}^{3}+\frac{5}{2}{x}^{2}+x+4$

Choice (a) is incorrect
Try again, the derivative of this function is ${F}^{\prime }\left(x\right)=-2sinx+3{x}^{2}+5x+1\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, the derivative of this function is ${F}^{\prime }\left(x\right)=2sinx+{x}^{2}+x+1\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, $F\left(0\right)=2$ but the derivative of this function is ${F}^{\prime }\left(x\right)=-2sinx+2{x}^{2}+5x+1\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is correct!
$\int 2sinx+2{x}^{2}+5x+1=2cosx+\frac{2}{3}{x}^{3}+\frac{5}{2}{x}^{2}+x+c\phantom{\rule{0.3em}{0ex}}.$
Since $F\left(0\right)=-2+0+0+0+0+c=2⇒\phantom{\rule{1em}{0ex}}c=4$ so the function
$F\left(x\right)=-2cosx+\frac{2}{3}{x}^{3}+\frac{5}{2}{x}^{2}+x+4$
A student wishes to find the area between the curve $8+2x-{x}^{2}$ and the $x$-axis and produces the following working. Which of the following statements are correct. Exactly one option must be correct)
 a) The limits of integration are wrong. b) The anti-derivative has been calculated incorrectly. c) The arithmetic is incorrect. d) The answer is correctly worked out at every step. e) None of the above.

Choice (a) is correct!
$8+2x-{x}^{2}=\left(4-x\right)\left(2+x\right)$ so the limits of integration should be -2 and 4. We then get a positive value for the area as expected.
Choice (b) is incorrect
Try again, the anti-derivative is correct.
Choice (c) is incorrect
Try again, the arithmetic is correct.
Choice (d) is incorrect
Try again, the answer is incorrect.
Choice (e) is incorrect
One of the above is the correct answer.
Which of the following integrals computes the area under the curve $y=\left(x+1\right)\left(x-1\right)\left(x-3\right)$? Exactly one option must be correct)
 a) ${\int }_{-1}^{1}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx+{\int }_{1}^{3}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx$ b) ${\int }_{-1}^{1}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx-{\int }_{-1}^{3}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx$ c) ${\int }_{-1}^{1}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx-{\int }_{1}^{3}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx$ d) ${\int }_{1}^{3}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx-{\int }_{-1}^{1}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx$

Choice (a) is incorrect
Try again, this gives ${\int }_{-1}^{3}{x}^{3}-3x-x+3\phantom{\rule{0.3em}{0ex}}dx$ which is not the required integral.
Choice (b) is incorrect
Try again, the limits of integration are incorrect.
Choice (c) is correct!
The curve looks like this

The value of the integral between -1 and 1 is positive and between 1 and 3 is negative so first we need to integrate between -1 and 1 and then, we need to integrate between 1 and 3 and change that integral’s sign. Thus we evaluate
${\int }_{-1}^{1}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx-{\int }_{1}^{3}{x}^{3}-3{x}^{2}-x+3\phantom{\rule{0.3em}{0ex}}dx$
Choice (d) is incorrect
Try again, you may need to sketch the curve.